我如何总结所有质数？

Question

I am working on an excercise to sum all prime numbers from 2 to the parameter. I have worked this far in the code, but am stuck. I believe by using the splice function, I am actually skipping an element because of a changed indices.

function sumPrimes(num) {
  var primearray = [];
  var sum = 0;
  for(var i =2; i <= num; i++){
    primearray.push(i);
  }

  for(var j = 0; j < primearray.length; j++) {
    console.log(primearray[j]);
    if ((primearray[j]%2===0) && (primearray[j] >2)) {
      primearray.splice(j,1);
    } else if ((primearray[j]%3===0) && (primearray[j] > 3)) {
      primearray.splice(j,1);
      console.log(primearray);
    } else if ((primearray[j]%5===0) && (primearray[j] > 5)) {
      primearray.splice(j,1);
    } else if ((primearray[j]%7===0) && (primearray[j] > 7)) {
      primearray.splice(j,1);
    }
  }
  sum = primearray.reduce();
  return sum;
}

sumPrimes(30);

I haven't utilized the reduce function yet because I am still working on the if else statements.

Answer 1

I found a pretty good solution to the same problem. afmeva was spot on. This is how it works.

function isPrime(val){

  //test if number is prime
  for(var i=2; i < val; i++){
    if(val % i === 0){
      return false;
    }
  }
  return true;
}

In the above code we accept a number to determine whether or not it is prime. We then loop from two all the way up until our number minus one because we know that our number will be divisible by itself and one. If the remainder of our value with the current loop value is zero then we know it is not prime so break out and say so.

This article explains very well

function sumPrimes(num) {
  var answer = 0;

  //loop through all numbers from 2 to input value

  for(var i=2; i <= num; i++){   

    //sum only prime numbers, skip all others
    if(isPrime(i)){
      answer += i;
    }
  }
  return answer;
}

sumPrimes(977); // 73156

Here's another good resource

Answer 2

 function sumPrimes(num) { let arr = Array.from({length: num+1}, (v, k) => k).slice(2); let onlyPrimes = arr.filter( (n) => { let m = n-1; while (m > 1 && m >= Math.sqrt(n)) { if ((n % m) === 0) return false; m--; } return true; }); return onlyPrimes.reduce((a,b) => a+b); } sumPrimes(977);

Answer 3

I have seen lots of people putting all prime numbers into arrays and in order to check if a number is prime, they check from 2 to the number to see if there's a remainder. You only need to check odd numbers, and only need to count to half the number because a number can't be divisible by any number greater than it has. Here's my solution:

function sumPrimes(num){
    var sum = num>=2?2:0;
    for(var i=3;i<=num;i+=2){
        var isPrime=true;
        for(var j=3;j<(i/2);j++){
            if (i%j==0)
            {
                isPrime=false;
                break;
            }
        }
        sum+=isPrime?i:0;
    }
    return sum;
}

Note: I started from j=2 because we are only checking odd numbers, so they'd never be divisible by 2.

Answer 4

function sumPrimes(num) {
  var sumArr= [];
  for(var i=0;i<=num;i++){
    if(isPrime(i))
      sumArr.push(i);
  }

  sumArr = sumArr.reduce(function(a,b){
    return a+b;
  })

  return sumArr;
}

function isPrime(num) {
    if(num < 2) return false;
    for (var i = 2; i < num; i++) {
        if(num%i === 0)
            return false;
    }
    return true;
}

sumPrimes(10);

Answer 5

something like this?

 function isPrime(_num) { for(var i = 2; i < _num; i++) { if(!(_num % i)) { return false } } return true; } function sumPrimes(_num) { var sum = 0; for(var i = 2; i <= _num; i++) { if(isPrime(i)) { sum += i; } } return sum; } sumPrimes(20) // 77 sumPrimes(5) // 10

Answer 6

You could do this as well.

 function sumPrimes(num) { var sum = 0; for (var i = 0; i <= num; i++) { if (isPrime(i)) { sum += i; } } return sum; } function isPrime(n) { if (n < 2) { return false; } if (n !== Math.round(n)) { return false; } var result = true; for (var i = 2; i <= Math.sqrt(n); i++) { if (n % i === 0) { result = false; } } return result; }

Answer 7

Here's my solution. I hope you find it easy to interpret:

function sumPrimes(num) {

  // determine if a number is prime
  function isPrime(n) {
    if (n === 2) return true;
    if (n === 3) return true;
    if (n % 2 === 0) return false;
    if (n % 3 === 0) return false;

    var  i = 5;
    var  w = 2;
    while (i * i <= n) {
        if (n % i === 0) {
            return false;
        }
        i += w;
        w = 6 - w;
    }
    return true;
  }

  // subtract 1 for 'not being prime' in my context
  var sum = isPrime(num) ? num - 1 : -1;

  for (var x = 0; x < num; x++) {
    if (isPrime(x) === true) {
      sum += x;
    }
  }

  return sum;
}

Answer 8

here is my solution to sum of n prime number

 function sumOfNPrimeNumber(num){ var sum = 0; const isPrime = function(n){ if (isNaN(n) || !isFinite(n) || n%1 || n<2) { return false; } if (n%2==0){ return (n==2); } var sqrt = Math.sqrt(n); for (var i = 3; i < sqrt; i+=2) { if(n%i == 0){ return false; } } return true; } const getNextPrime = function* (){ let nextNumber = 2; while(true){ if(isPrime(nextNumber)){ yield nextNumber; } ++nextNumber; } } const nextPrime = getNextPrime(); for (var i = 0; i < num; i++) { sum = sum + nextPrime.next().value; } return sum; } console.log(sumOfNPrimeNumber(3));

Answer 9

All the above answers make use of helper functions or aren't time efficients. This is a quick, recursive solution in O(n) time:

// @ signature int -> int
// @ interpr: returns sum of all prime integers <= num
// assume: num is positive integer

function sumPrimes(num) {
  if (num <= 2) {
    return 2;
  }

  let i = 2;
  while (i < num) {
    if (num % i === 0) {
      return sumPrimes(num - 1)
    }
    i++;
  }
  
  return num + sumPrimes(num - 1)
}

// test
sumPrimes(10); // -> 17

Answer 10

The following solution uses the Eratosthenes Sieve to sum all prime numbers lower than or equal to num . The first for loop fills an array with size equal to num with true . The second for loop sets to false all non-prime numbers in the array . Then, the last for loop simply iterates through the array to sum all the array indexes i for which the value in the array, ie, array[i] , is equal to true .

/**
 * Sum all primes lower or equal than n.
 * Uses the Eratosthenes Sieve to find all primes under n.
 */
function sumPrimes(num) {
  let array = [];
  let output = 0;

  // Fill an array of boolean with 'true' from 2 to n.
  for (let i = 0; i <= num; i++) {
    array.push(true);
  }

  // Set all multiples of primes to 'false' in the array.
  for (let i = 2; i <= Math.sqrt(num); i++) {
    if (array[i]) {
      for (let j = i * i; j <= num; j += i) {
        array[j] = false;
      }
    }
  }

  // All array[i] set to 'true' are primes, so we just need to add them all.
  for (var i = 2; i <= num; i++) {
    if (array[i]) {
      output += i;
    }
  }

  return output;
}

console.log(sumPrimes(10)); // 17
console.log(sumPrimes(977)); // 73156
console.log(sumPrimes(250_000_000)); // 197558914577

Answer 11

This is what I've done to get primes. I don't know if it's the most efficient, but it works. This is in Java, but can be easily converted to JavaScript. Hopefully this will help point you in the right direction.

final int TOTAL = 10;
int primes[] = new int[TOTAL];
int arrPos = 2;
boolean prime = false;
primes[0] = 2;

for (int i = 2; i < TOTAL; i++) {
  prime = false;
  int sqrt = (int) Math.sqrt(i);
  for (int j = 1; j < arrPos && primes[j] < sqrt; j++) {
    if (i % primes[j] != 0) {
      prime = true;
    } else {
      prime = false;
      break;
    }
  }

  if (prime == true) {
    primes[arrPos] = i;
    arrPos++;
  }
}

Answer 12

function prime_sum(num){
let count=0;        *//tracks the no of times number is divided perfectly*
   for(let i=1;i<=num;i++){        *//from 1 upto the number*
      if(num%i==0){count++};
}
if(count===2){return "Prime"};
return{"Not prime"};
}
console.log(prime_sum(10));//expected output is 17**

//the code receives a number,checks through the range and returns prime if it meets the condition